1. **State the problem:** We want to find the group size $x$ (where $x \geq 25$) that maximizes the revenue for a travel agency. The price per person decreases by 10 for each person above 25, starting at 300 for exactly 25 people. 2. **Define variables and formula:** Let $x$ be the number of people in the group. Price per person when group size is $x$: $$p(x) = 300 - 10(x - 25)$$ Revenue $R(x)$ is the number of people times the price per person: $$R(x) = x \times p(x) = x \times \bigl(300 - 10(x - 25)\bigr)$$ 3. **Simplify the revenue function:** $$R(x) = x \times (300 - 10x + 250) = x \times (550 - 10x)$$ $$R(x) = 550x - 10x^2$$ 4. **Find the vertex of the parabola $R(x) = -10x^2 + 550x$:** The parabola opens downward (coefficient of $x^2$ is negative), so the vertex gives the maximum revenue. Vertex $x$-coordinate formula: $$x = -\frac{b}{2a} = -\frac{550}{2 \times (-10)} = \frac{550}{20} = 27.5$$ 5. **Calculate maximum revenue at $x=27.5$:** $$R(27.5) = 550 \times 27.5 - 10 \times (27.5)^2$$ $$= 15125 - 10 \times 756.25 = 15125 - 7562.5 = 7562.5$$ 6. **Interpretation:** The maximum revenue is 7562.5 when the group size is 27.5 people. Since group size must be an integer, check $x=27$ and $x=28$: $$R(27) = 550 \times 27 - 10 \times 27^2 = 14850 - 7290 = 7560$$ $$R(28) = 550 \times 28 - 10 \times 28^2 = 15400 - 7840 = 7560$$ Both 27 and 28 people produce maximum revenue of 7560. 7. **Find intercepts (where revenue is zero):** Set $R(x) = 0$: $$550x - 10x^2 = 0$$ $$10x(55 - x) = 0$$ So, $$x=0 \quad \text{or} \quad x=55$$ Since group size must be at least 25, the relevant intercepts are at $x=25$ (start) and $x=55$ (where revenue returns to zero). **Final answer:** The group size that produces maximum revenue is 27 or 28 people, with maximum revenue 7560.